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# Check If N and Its Double Exist

## [Check If N and Its Double Exist](https://leetcode.com/problems/check-if-n-and-its-double-exist)

Given an array `arr` of integers, check if there exists two integers `N` and `M` such that `N` is the double of `M` ( i.e. `N = 2 * M`).

More formally check if there exists two indices `i` and `j` such that :

* `i != j`
* `0 <= i, j < arr.length`
* `arr[i] == 2 * arr[j]`

**Example 1:**

```

Input: arr = [10,2,5,3]
Output: true
Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.
```

**Example 2:**

```

Input: arr = [7,1,14,11]
Output: true
Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.
```

**Example 3:**

```

Input: arr = [3,1,7,11]
Output: false
Explanation: In this case does not exist N and M, such that N = 2 * M.
```

**Constraints:**

* `2 <= arr.length <= 500`
* `-10^3 <= arr[i] <= 10^3`

## Solutions

### 🧠 Cpp

```cpp
#define all(x) (x).begin(), (x).end()

class Solution
{
public:
    bool checkIfExist(vector<int>& arr)
    {
        //O(n^2) solution
        for(auto iter = begin(arr); iter < end(arr); ++iter)
        {
            auto found = find(all(arr), *iter*2);
            if( found != arr.end() && found != iter)
                return true;
        }

        return false;
    }
};
```